The problem is as follows:
Let $ \Sigma_3=\left\{\begin{bmatrix}0\\0\\0\end{bmatrix},\begin{bmatrix}0\\0\\1\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix},\ldots,\begin{bmatrix}1\\1\\1\end{bmatrix}\right\} $
So $\Sigma_3$ contains all size 3 columns of 0s and 1s. A string of symbols in $\Sigma_3$ give three rows of 0s and 1s. Consider each row to be a binary number, and let
$$B = \left\{ w \in (\Sigma_3)^* \mid \text{ the bottom row of } w \text{ is the sum of the top two rows} \right\}$$
So $\begin{bmatrix}0\\0\\1\end{bmatrix}\begin{bmatrix}1\\0\\0\end{bmatrix}\begin{bmatrix}1\\1\\0\end{bmatrix} \in B$, but $\begin{bmatrix}0\\0\\1\end{bmatrix}\begin{bmatrix}1\\0\\1\end{bmatrix} \notin B$
Show that $B$ is regular.
Now, there's a hint in there that tells us we can try showing that $B^R$ is regular, where $B^R$ is the reverse of $B$. (It is assumed that the reverse of a regular language is also regular, shown in a separate problem).
Now, here's my solution:
Let $a = \begin{bmatrix}1\\0\\1\end{bmatrix}, b = \begin{bmatrix}1\\1\\0\end{bmatrix}, c = \begin{bmatrix}0\\1\\1\end{bmatrix}, d = \begin{bmatrix}0\\0\\0\end{bmatrix}$
And then let
$e = \begin{bmatrix}1\\0\\0\end{bmatrix}, f = \begin{bmatrix}0\\1\\0\end{bmatrix}, g = \begin{bmatrix}1\\1\\1\end{bmatrix}, h = \begin{bmatrix}0\\0\\1\end{bmatrix}$
We'd get the following finite state machine recognizing $B^R$:
Basically, $q_2$ is the "carry" state, whereas $q_1$ is the final state as there is no carrying necessary.
Since we're solving this in a reverse fashion, i.e. solving for $B^R$, there are only four valid first choices out of 8 in total: $a$, $c$, and $d$, for which there is no carrying, and $b$ for which there is.
If we're in state $q_2$, which means we're carrying the 1, only $e$, $f$, $g$ and $h$ make sense. Out of those, only $h$ can send us to $q_1$, since there's nothing to carry.
I apologize if my English makes the above sounds awkward (especially the "carrying" part), but I hope the solution is understandable.